MATHEMATICS II - Deneme Sınavı - 10

which means the right answer is C.

Therefore, we evaluate the indefinite integral as

Hence we obtain

The right answer is C.

Hence, we have to evaluate the definite integral

on the interval

Therefore required area  is evaluated as

that means the right answer is A.

Hence the right answer is D.

Hence the right answer is C.

required area  is evaluated with the definite integral

and the limits of integration are changed from 0 to 4. Therefore, we have

The right answer is E.

and

The area of the region corresponding to the interval

equals to the area corresponding to

Hence required area  is calculated by the definite integral

Because we observe that

and

Thus, we get

The right answer is B.

is denoted by

then the average value is calculated by the definite integral

For

integral becomes

Therefore, the correct answer is A.

on the interval

and since the function is even the area A can be calculated by definite integral:

Calculating the integral gives the area as

Hence the correct answer is C.

and the upper bound becomes

and the lower bound becomes

Substituting these values into the integral we get

which gives the correct answer B.

So the correct answer is D.

we have

Hence, the area can be calculated with the formula

Hence the right answer is D.

which gives

Also, let

which gives

Substituting these values into

we get;

Therefore, the correct answer is C.

then

also the boundaries become

and

Substituting these values into the question we have

Hence the correct answer is E.

the definite integral can be written as

Considering this we can write the question as

So;

So the correct answer is A.

The correct answer is E.

let us divide the interval

into four subintervals

and

Since the function f is an increasing function, it takes its smallest values at the left end of each interval. The length of each subinterval is

and

therefore

let us divide the interval

into four subintervals

and

Since the function f is an increasing function, it takes its smallest values at the left end of each interval. The length of each subinterval is

and

therefore

Thus lower sum of  corresponding to given partition is

Since

and since

Thus upper sum of  corresponding to the given partition is